[Ybt] [Basic calculation recurrence class example 4] Passing game

Passing game

Topic link: passing game

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Title description

Problem solving ideas

This question is relatively simple.
Obviously a person can get the ball from both sides of him, then it is also possible to get contributions from both sides.
The recurrence formula is:
$$f_{i,j}=f_{i-1,j-1}+f_{i-1,j+1}$$
Give $f_{0,1}$Attached 1 1The initial value of$1$ ,$j=n$ or$j=A$ special sentence is sufficient at$0:00$ .

code

#include<iostream>
#include<cstdio>
#define int long long
using namespace std;

int n,m;
int f[40][40];

signed main()
{
    
    
	cin>>n>>m;
	f[0][1]=1;
	for(int i=1;i<=m;i++)
		for(int j=1;j<=n;j++)
		{
    
    
			if(j==1)
				f[i][j]=f[i-1][n]+f[i-1][2];
			else if(j==n)
				f[i][j]=f[i-1][n-1]+f[i-1][1];
			else
				f[i][j]=f[i-1][j+1]+f[i-1][j-1];
		}
	cout<<f[m][1]<<endl;
}